🧠:<think>First, I need to determine the total area of the store that needs to be covered by the sprinkler system. The store is rectangular, so I'll calculate the area by multiplying its length by its width. The length is 50 meters and the width is 30 meters, which gives a total area of 1500 square meters.Next, each sprinkler head covers an area of 25 square meters. To find out how many sprinkler heads are needed, I'll divide the total area by the coverage area of one sprinkler. Dividing 1500 square meters by 25 square meters per sprinkler head results in 60 sprinkler heads required.Now, to calculate the total water flow rate for the entire system, I'll multiply the number of sprinkler heads by the flow rate per head. Each sprinkler head has a flow rate of 0.2 liters per second, so multiplying 60 by 0.2 liters per second gives a total flow rate of 12 liters per second.For the water pressure calculation, the formula provided is ( P = k cdot sqrt{h} ), where ( k ) is 1000 and ( h ) is the height of the sprinkler heads. The store's height is 4 meters, so I'll take the square root of 4, which is 2. Multiplying 1000 by 2 results in a required water pressure of 2000 pascals.</think>

🧠:<think>Okay, so I have this problem where a policy advisor and a researcher are working on electoral reforms. They've come up with a probability density function to model voter turnout based on the number of hours spent on voter outreach. The function is given as ( f(x) = kxe^{-x/2} ), where ( x ) is the number of hours, and ( k ) is a normalization constant. There are two parts to this problem. The first part is to find the value of ( k ) that normalizes the probability density function over the interval from 0 to infinity. The second part is to calculate the probability that voter turnout will be maximized within a budget constraint of 10 hours. Starting with the first part, I remember that for a function to be a valid probability density function (pdf), the integral of the function over its entire domain must equal 1. So, I need to integrate ( f(x) ) from 0 to infinity and set that equal to 1 to solve for ( k ). The function given is ( f(x) = kxe^{-x/2} ). So, the integral we need to compute is:[int_{0}^{infty} kxe^{-x/2} dx = 1]I can factor out the constant ( k ) from the integral:[k int_{0}^{infty} xe^{-x/2} dx = 1]Now, I need to compute the integral ( int_{0}^{infty} xe^{-x/2} dx ). This seems like a standard integral that can be solved using integration by parts. Let me recall the formula for integration by parts:[int u dv = uv - int v du]Let me set ( u = x ) and ( dv = e^{-x/2} dx ). Then, ( du = dx ) and ( v = int e^{-x/2} dx ). To find ( v ), I need to integrate ( e^{-x/2} ). The integral of ( e^{ax} ) is ( frac{1}{a}e^{ax} ), so in this case, ( a = -1/2 ). Therefore,[v = int e^{-x/2} dx = -2e^{-x/2}]Now, applying integration by parts:[int xe^{-x/2} dx = uv - int v du = x(-2e^{-x/2}) - int (-2e^{-x/2}) dx]Simplify this:[= -2xe^{-x/2} + 2 int e^{-x/2} dx]We already know that ( int e^{-x/2} dx = -2e^{-x/2} ), so substituting back:[= -2xe^{-x/2} + 2(-2e^{-x/2}) + C][= -2xe^{-x/2} - 4e^{-x/2} + C]Now, we need to evaluate this from 0 to infinity. Let's compute the limit as ( x ) approaches infinity and subtract the value at ( x = 0 ).First, as ( x ) approaches infinity, let's analyze the terms:- ( -2xe^{-x/2} ): The exponential term ( e^{-x/2} ) decays to 0 faster than the polynomial term ( x ) grows, so this term goes to 0.- ( -4e^{-x/2} ): Similarly, this term also goes to 0 as ( x ) approaches infinity.So, the upper limit evaluates to 0.Now, evaluating at ( x = 0 ):- ( -2(0)e^{-0/2} = 0 )- ( -4e^{-0/2} = -4(1) = -4 )Therefore, the integral from 0 to infinity is:[[0] - [0 - 4] = 0 - (-4) = 4]So, the integral ( int_{0}^{infty} xe^{-x/2} dx = 4 ).Going back to our equation:[k times 4 = 1][k = frac{1}{4}]So, the normalization constant ( k ) is ( 1/4 ).Now, moving on to the second part of the problem. We need to calculate the probability that the voter turnout will be maximized within a budget constraint of 10 hours. Wait, the wording here is a bit confusing. It says, "the probability that the voter turnout will be maximized within this budget constraint." Hmm. So, does this mean we need to find the probability that the maximum voter turnout occurs within 10 hours? Or perhaps, given that the maximum occurs at some point, what's the probability that it's within 10 hours? Alternatively, maybe it's asking for the probability that the number of hours spent on outreach is less than or equal to 10, given that the function is defined over [0, ∞). But the way it's phrased, "the probability that the voter turnout will be maximized within this budget constraint," makes me think it's about the maximum of the function.Wait, actually, the function ( f(x) = kxe^{-x/2} ) is a probability density function for the number of hours ( x ) spent on outreach. So, the voter turnout is modeled by this pdf, and we need to find the probability that ( x ) is less than or equal to 10. That is, the probability that the number of hours spent is within the budget constraint of 10 hours.Alternatively, maybe it's about the maximum value of the pdf. Let me think.Wait, the pdf is ( f(x) = (1/4)xe^{-x/2} ). So, it's a gamma distribution, right? Because the gamma distribution has the form ( f(x) = frac{1}{Gamma(k)theta^k}x^{k-1}e^{-x/theta} ). In this case, comparing to our function, ( k = 2 ) and ( theta = 2 ), since ( f(x) = frac{1}{4}xe^{-x/2} ). So, it's a gamma distribution with shape parameter 2 and scale parameter 2.In a gamma distribution, the mean is ( ktheta ), which would be 4 in this case. The mode is ( (k - 1)theta ), which is ( (2 - 1) times 2 = 2 ). So, the mode is at 2 hours. That is, the pdf reaches its maximum at 2 hours.But the question is about the probability that the voter turnout will be maximized within a budget constraint of 10 hours. Hmm, maybe I misinterpreted the question. Perhaps it's asking for the probability that the maximum voter turnout occurs within 10 hours. But since the maximum occurs at 2 hours, which is well within 10 hours, the probability would be 1. That seems too straightforward.Alternatively, perhaps the question is asking for the probability that the number of hours ( x ) is less than or equal to 10, given that the pdf is defined over [0, ∞). So, we need to compute the cumulative distribution function (CDF) at 10, which is the integral of ( f(x) ) from 0 to 10.Yes, that makes more sense. So, the probability that ( x leq 10 ) is the integral from 0 to 10 of ( f(x) dx ). Since ( f(x) ) is a pdf, this integral will give the probability that the number of hours is within the budget.So, let's compute:[P(X leq 10) = int_{0}^{10} frac{1}{4}xe^{-x/2} dx]We can factor out the 1/4:[= frac{1}{4} int_{0}^{10} xe^{-x/2} dx]We already computed the integral ( int xe^{-x/2} dx ) earlier, which was 4 when evaluated from 0 to infinity. Now, we need to compute it from 0 to 10.Using the same integration by parts method as before, let's compute the indefinite integral first.Let me recall that earlier, we found:[int xe^{-x/2} dx = -2xe^{-x/2} - 4e^{-x/2} + C]So, the definite integral from 0 to 10 is:[[-2xe^{-x/2} - 4e^{-x/2}]_{0}^{10}]Let's compute this at 10 and at 0.First, at ( x = 10 ):[-2(10)e^{-10/2} - 4e^{-10/2} = -20e^{-5} - 4e^{-5} = (-20 - 4)e^{-5} = -24e^{-5}]Now, at ( x = 0 ):[-2(0)e^{-0/2} - 4e^{-0/2} = 0 - 4(1) = -4]So, subtracting the lower limit from the upper limit:[[-24e^{-5}] - [-4] = -24e^{-5} + 4 = 4 - 24e^{-5}]Therefore, the integral ( int_{0}^{10} xe^{-x/2} dx = 4 - 24e^{-5} ).Now, multiplying by 1/4:[P(X leq 10) = frac{1}{4}(4 - 24e^{-5}) = 1 - 6e^{-5}]So, the probability is ( 1 - 6e^{-5} ).Let me compute this numerically to get a sense of the value.First, compute ( e^{-5} ). ( e^{-5} ) is approximately ( 0.006737947 ).So, ( 6e^{-5} approx 6 times 0.006737947 approx 0.04042768 ).Therefore, ( 1 - 0.04042768 approx 0.95957232 ).So, approximately 95.96% probability.Wait, that seems quite high. Let me double-check my calculations.First, the integral from 0 to 10 of ( xe^{-x/2} dx ) was computed as 4 - 24e^{-5}. Let me verify that.Wait, when I computed the definite integral earlier, I had:At x=10: -24e^{-5}At x=0: -4So, the definite integral is (-24e^{-5}) - (-4) = -24e^{-5} + 4.Yes, that's correct.Then, multiplying by 1/4 gives 1 - 6e^{-5}, which is approximately 1 - 0.0404 = 0.9596.So, about 95.96% probability.That seems correct because the gamma distribution with shape 2 and scale 2 has a mean of 4, so 10 is quite far in the tail, but since the distribution is skewed, the probability of being less than 10 is high.Alternatively, perhaps I made a mistake in setting up the integral. Let me think again.Wait, the pdf is ( f(x) = frac{1}{4}xe^{-x/2} ). So, integrating from 0 to 10, we get the CDF at 10, which is the probability that ( X leq 10 ).Yes, that's correct. So, the calculation seems right.Therefore, the probability is ( 1 - 6e^{-5} ), which is approximately 0.9596.So, summarizing:1. The normalization constant ( k ) is ( 1/4 ).2. The probability that the number of hours spent on outreach is within the budget constraint of 10 hours is ( 1 - 6e^{-5} ), approximately 95.96%.I think that's it. Let me just make sure I didn't make any arithmetic errors.Wait, when I computed the integral from 0 to 10, I had:[[-2xe^{-x/2} - 4e^{-x/2}]_{0}^{10} = (-20e^{-5} -4e^{-5}) - (0 -4) = (-24e^{-5}) - (-4) = -24e^{-5} +4]Yes, that's correct.Then, multiplying by 1/4:[frac{1}{4}(4 -24e^{-5}) = 1 -6e^{-5}]Yes, that's correct.So, all steps seem correct. Therefore, the answers are:1. ( k = 1/4 )2. Probability ( = 1 -6e^{-5} approx 0.9596 )</think>

🧠:<think>Okay, so I have this problem about a geologist analyzing a region for a new bridge. There are two sub-problems here. Let me tackle them one by one.Starting with Sub-problem 1. The geologist has identified three layers: surface soil, middle rock, and deep groundwater-saturated layer. Each has different thicknesses and Young's moduli. I need to calculate the effective Young's modulus using the rule of mixtures for series and parallel configurations. Hmm, I remember that the rule of mixtures is used to estimate the effective properties of composite materials. But wait, in this case, the layers are stacked on top of each other, so I think they are in series. Or is it parallel? Let me think.In composite materials, when materials are arranged in series, meaning the same load is applied to each, the compliance (which is the inverse of Young's modulus) adds up. So, the effective modulus would be the reciprocal of the sum of the reciprocals of each layer's modulus, weighted by their thickness. On the other hand, if they were in parallel, meaning the same strain is applied, then the moduli would add up. But in this case, the layers are stacked, so they are in series. So, I think I should use the series rule of mixtures.Let me write down the formula. For series configuration, the effective Young's modulus ( E_{text{eff}} ) is given by:[frac{1}{E_{text{eff}}} = frac{t_1}{E_1} + frac{t_2}{E_2} + frac{t_3}{E_3}]Where ( t_1, t_2, t_3 ) are the thicknesses of each layer, and ( E_1, E_2, E_3 ) are their respective Young's moduli.Given:- Surface soil: ( t_1 = 15 ) m, ( E_1 = 20 ) MPa- Middle rock: ( t_2 = 25 ) m, ( E_2 = 50 ) MPa- Groundwater-saturated: ( t_3 = 20 ) m, ( E_3 = 10 ) MPaWait, but I need to make sure the units are consistent. All thicknesses are in meters, and moduli are in MPa, which is consistent.So, plugging in the numbers:First, calculate each term ( frac{t_i}{E_i} ):For the surface soil:[frac{15}{20} = 0.75 text{ m·MPa}^{-1}]Middle rock:[frac{25}{50} = 0.5 text{ m·MPa}^{-1}]Groundwater-saturated:[frac{20}{10} = 2 text{ m·MPa}^{-1}]Adding them up:[0.75 + 0.5 + 2 = 3.25 text{ m·MPa}^{-1}]Therefore, the effective Young's modulus is:[E_{text{eff}} = frac{1}{3.25} approx 0.3077 text{ MPa}]Wait, that seems really low. Is that right? Let me double-check. The surface soil has a low modulus, and the groundwater-saturated layer also has a low modulus, so their contributions would be higher in the reciprocal space. So, the effective modulus being low makes sense because the structure is dominated by the softer layers.But let me think again. The formula is correct for series, right? Because each layer is in series, so the strain is the same, but the stresses add up. Wait, no, actually, in series, the strain is the same, but the stress is different. Hmm, maybe I confused series and parallel.Wait, no, in composite materials, when materials are in series, meaning the same strain is applied, the stress is different. So, the effective modulus is the sum of the moduli multiplied by their respective strain. Wait, I'm getting confused.Let me recall: For materials in series (same strain), the total stress is the sum of the stresses in each material. Since stress is modulus times strain, the total stress is ( E_1 epsilon + E_2 epsilon + E_3 epsilon = (E_1 + E_2 + E_3) epsilon ). Therefore, the effective modulus would be ( E_{text{eff}} = E_1 + E_2 + E_3 ). But that doesn't make sense because adding moduli would give a higher modulus, but in reality, the effective modulus should be lower if the materials are in series because the softer layers would dominate.Wait, no, actually, if you have materials in series, the effective modulus is given by the harmonic mean weighted by thickness. So, the formula I used earlier is correct.But let me think about units. The units of ( t/E ) is meters per MPa. When you add them up, you get meters per MPa. Then, the reciprocal is MPa per meter. But modulus is in MPa, so that doesn't make sense. Wait, maybe I messed up the formula.Alternatively, perhaps the effective modulus is calculated as the sum of each layer's modulus multiplied by their thickness, divided by the total thickness. Wait, that would be the parallel case.Wait, no, I think I need to clarify. In composite materials, when the materials are in series (same strain), the effective modulus is given by:[frac{1}{E_{text{eff}}} = frac{t_1}{E_1} + frac{t_2}{E_2} + frac{t_3}{E_3}]Which is what I did earlier. So, the units would be (m / MPa) summed up, and then reciprocal gives MPa. Wait, no, reciprocal of (m / MPa) is MPa / m, which is not modulus. Hmm, something is wrong here.Wait, maybe I need to consider the reciprocal correctly. Let me think in terms of compliance. Compliance ( C ) is ( 1/E ). For series, the total compliance is the sum of the compliances of each layer. But compliance is in m/MPa, so if I sum them up, I get total compliance in m/MPa. Then, the effective modulus is ( 1 / C_{text{total}} ), which would be in MPa/m? That doesn't make sense.Wait, no, I think I need to consider the total strain. Let me think differently. Suppose a load is applied, causing a strain ( epsilon ) in each layer. The total strain is the same for each layer because they are in series. The total stress is the sum of the stresses in each layer. So, ( sigma = sigma_1 + sigma_2 + sigma_3 = E_1 epsilon + E_2 epsilon + E_3 epsilon = (E_1 + E_2 + E_3) epsilon ). Therefore, the effective modulus is ( E_{text{eff}} = sigma / epsilon = E_1 + E_2 + E_3 ).But that would mean the effective modulus is just the sum of the individual moduli, which seems too simplistic. But in this case, the moduli are 20, 50, and 10 MPa, so the sum would be 80 MPa. That seems high, but maybe that's correct?Wait, no, because the layers have different thicknesses. So, the contribution of each layer to the total stress isn't just additive because the strain is the same, but the stress in each layer is modulus times strain, so the total stress is the sum of each modulus times strain. Therefore, the effective modulus is the sum of the moduli. But that doesn't account for thickness.Wait, now I'm really confused. Maybe I need to think about it in terms of stiffness. The stiffness of a layer is ( k = E cdot A / t ), where A is the area. But since all layers have the same area, the total stiffness in series would be ( 1/(1/k_1 + 1/k_2 + 1/k_3) ). But that seems complicated.Alternatively, perhaps the effective modulus is given by the weighted harmonic mean based on thickness. So, the formula is:[E_{text{eff}} = frac{t_1 + t_2 + t_3}{frac{t_1}{E_1} + frac{t_2}{E_2} + frac{t_3}{E_3}}]Yes, that makes sense. Because if all layers have the same thickness, it's the harmonic mean. If one layer is thicker, it contributes more to the effective modulus. So, the formula is:[E_{text{eff}} = frac{t_1 + t_2 + t_3}{frac{t_1}{E_1} + frac{t_2}{E_2} + frac{t_3}{E_3}}]So, let me recalculate using this formula.First, total thickness ( T = 15 + 25 + 20 = 60 ) meters.Then, the denominator is ( frac{15}{20} + frac{25}{50} + frac{20}{10} = 0.75 + 0.5 + 2 = 3.25 ) m·MPa^{-1}So, ( E_{text{eff}} = frac{60}{3.25} approx 18.46 ) MPa.That seems more reasonable. So, the effective modulus is approximately 18.46 MPa.Wait, but earlier I thought the formula was reciprocal, but now I see that the formula is total thickness divided by the sum of ( t_i / E_i ). So, that makes sense because it's the harmonic mean weighted by thickness.So, I think this is the correct approach.Now, moving on to Sub-problem 2. The geologist needs to determine the critical cyclic stress ratio (CSR) using the Seed and Idriss method. The formula given is:[CSR = 0.65 times frac{sigma_v'}{sigma_v}]Where ( sigma_v' ) is the effective vertical stress and ( sigma_v ) is the total vertical stress. The total vertical stress at the top of the groundwater-saturated layer is 600 kPa, and the pore water pressure is 300 kPa. The threshold CSR for liquefaction is 0.3. We need to calculate CSR and determine if the layer is at risk.First, I need to find ( sigma_v' ) and ( sigma_v ).Given that the total vertical stress ( sigma_v ) is 600 kPa. The pore water pressure ( u ) is 300 kPa. Therefore, the effective vertical stress ( sigma_v' = sigma_v - u = 600 - 300 = 300 ) kPa.So, plugging into the CSR formula:[CSR = 0.65 times frac{300}{600} = 0.65 times 0.5 = 0.325]So, the CSR is 0.325. The threshold is 0.3, so since 0.325 > 0.3, the layer is at risk of liquefaction.Wait, but let me make sure I didn't make a mistake. The formula is correct? Yes, it's 0.65 times the ratio of effective to total stress. The effective stress is total minus pore pressure, which is 300 kPa. So, 300/600 is 0.5. 0.65 * 0.5 is indeed 0.325. So, yes, it's above the threshold.Therefore, the layer is at risk of liquefaction.So, summarizing:Sub-problem 1: Effective Young's modulus is approximately 18.46 MPa.Sub-problem 2: CSR is 0.325, which is above the threshold, so liquefaction risk exists.Final AnswerSub-problem 1: The effective Young's modulus is boxed{18.46 text{ MPa}}.Sub-problem 2: The critical cyclic stress ratio is boxed{0.325}, indicating the layer is at risk of liquefaction.</think>

🧠:<think>Okay, so I have these two differential equation problems to solve, and they're about an athlete named Alex preparing for a comeback. The first one is about endurance, modeled by a logistic growth equation, and the second one is about speed, modeled by another differential equation. Let me try to tackle them one by one.Starting with problem 1: The logistic growth model for endurance. The differential equation is given as:[ frac{dE}{dt} = rEleft(1 - frac{E}{K}right) ]They give me the initial condition ( E(0) = 1.5 ) hours, the growth rate ( r = 0.2 ), and the carrying capacity ( K = 5 ) hours. I need to find the expression for ( E(t) ) and then determine the time ( t ) it takes for Alex to reach 4 hours of endurance.Alright, so I remember that the logistic equation is a common model for population growth, but here it's applied to endurance improvement. The standard solution to the logistic differential equation is:[ E(t) = frac{K}{1 + left(frac{K - E(0)}{E(0)}right)e^{-rt}} ]Let me verify that. Yes, that's the general solution for the logistic equation. So plugging in the given values:( E(0) = 1.5 ), ( K = 5 ), ( r = 0.2 ).First, compute ( frac{K - E(0)}{E(0)} ):( frac{5 - 1.5}{1.5} = frac{3.5}{1.5} = frac{7}{3} approx 2.333 ).So the expression becomes:[ E(t) = frac{5}{1 + frac{7}{3}e^{-0.2t}} ]Simplify that:[ E(t) = frac{5}{1 + frac{7}{3}e^{-0.2t}} ]I can write it as:[ E(t) = frac{5}{1 + frac{7}{3}e^{-0.2t}} ]Alternatively, to make it look cleaner, I can write 7/3 as a decimal, but maybe it's better to keep it as a fraction for exactness.Now, to find the time ( t ) when ( E(t) = 4 ) hours.Set up the equation:[ 4 = frac{5}{1 + frac{7}{3}e^{-0.2t}} ]Solve for ( t ).First, multiply both sides by the denominator:[ 4left(1 + frac{7}{3}e^{-0.2t}right) = 5 ]Divide both sides by 4:[ 1 + frac{7}{3}e^{-0.2t} = frac{5}{4} ]Subtract 1 from both sides:[ frac{7}{3}e^{-0.2t} = frac{5}{4} - 1 = frac{1}{4} ]Multiply both sides by ( frac{3}{7} ):[ e^{-0.2t} = frac{1}{4} times frac{3}{7} = frac{3}{28} ]Take the natural logarithm of both sides:[ -0.2t = lnleft(frac{3}{28}right) ]Solve for ( t ):[ t = frac{lnleft(frac{3}{28}right)}{-0.2} ]Compute the value:First, calculate ( ln(3/28) ). Let me compute that:( ln(3) approx 1.0986 ), ( ln(28) approx 3.3322 ), so ( ln(3/28) = ln(3) - ln(28) approx 1.0986 - 3.3322 = -2.2336 ).So,[ t = frac{-2.2336}{-0.2} = frac{2.2336}{0.2} = 11.168 ]So approximately 11.168 weeks. Since the question asks for the time in weeks, I can round it to two decimal places, so 11.17 weeks.Wait, let me double-check my calculations.Compute ( ln(3/28) ):3 divided by 28 is approximately 0.107142857.Then, ( ln(0.107142857) approx -2.2336 ). Yes, that's correct.So, ( t = (-2.2336)/(-0.2) = 11.168 ). So, 11.168 weeks, which is approximately 11.17 weeks.Alternatively, if they want an exact expression, it's ( t = frac{ln(28/3)}{0.2} ), but since 28/3 is approximately 9.3333, ( ln(9.3333) approx 2.2336 ), so same result.So, that's problem 1 done.Moving on to problem 2: The differential equation for speed is:[ frac{dS}{dt} = aS - bS^2 ]Given ( S(0) = 6 ) mph, ( a = 0.1 ), ( b = 0.01 ). Need to solve the differential equation and determine the long-term behavior as ( t ) approaches infinity.Alright, so this is a Bernoulli equation, but it's also a logistic-type equation, similar to the first problem. Let me write it as:[ frac{dS}{dt} = aS - bS^2 = S(a - bS) ]So, it's a separable equation. Let's try to solve it.Separate variables:[ frac{dS}{S(a - bS)} = dt ]Integrate both sides.First, let's write the left side as partial fractions.Let me set up:[ frac{1}{S(a - bS)} = frac{A}{S} + frac{B}{a - bS} ]Multiply both sides by ( S(a - bS) ):[ 1 = A(a - bS) + B S ]Expand:[ 1 = Aa - AbS + BS ]Group terms:[ 1 = Aa + S(-Ab + B) ]Since this must hold for all S, the coefficients of like terms must be equal on both sides.So, for the constant term: ( Aa = 1 ) => ( A = 1/a ).For the coefficient of S: ( -Ab + B = 0 ) => ( B = Ab = (1/a)b = b/a ).Therefore, the partial fractions decomposition is:[ frac{1}{S(a - bS)} = frac{1}{a S} + frac{b}{a(a - bS)} ]So, the integral becomes:[ int left( frac{1}{a S} + frac{b}{a(a - bS)} right) dS = int dt ]Compute the integrals:Left side:First term: ( frac{1}{a} int frac{1}{S} dS = frac{1}{a} ln|S| )Second term: ( frac{b}{a} int frac{1}{a - bS} dS )Let me make a substitution for the second integral: Let ( u = a - bS ), then ( du = -b dS ), so ( -du/b = dS ).Therefore, the second integral becomes:( frac{b}{a} times left( -frac{1}{b} int frac{1}{u} du right) = -frac{1}{a} ln|u| + C = -frac{1}{a} ln|a - bS| + C )So, combining both terms:Left side integral:[ frac{1}{a} ln|S| - frac{1}{a} ln|a - bS| + C = frac{1}{a} lnleft|frac{S}{a - bS}right| + C ]Right side integral:[ int dt = t + C ]So, putting it together:[ frac{1}{a} lnleft|frac{S}{a - bS}right| = t + C ]Multiply both sides by ( a ):[ lnleft|frac{S}{a - bS}right| = a t + C ]Exponentiate both sides:[ left|frac{S}{a - bS}right| = e^{a t + C} = e^{C} e^{a t} ]Let me write ( e^{C} ) as another constant ( C' ), since it's just a positive constant.So,[ frac{S}{a - bS} = C' e^{a t} ]Now, solve for S.Multiply both sides by ( a - bS ):[ S = C' e^{a t} (a - bS) ]Expand:[ S = a C' e^{a t} - b C' e^{a t} S ]Bring the term with S to the left:[ S + b C' e^{a t} S = a C' e^{a t} ]Factor out S:[ S(1 + b C' e^{a t}) = a C' e^{a t} ]Solve for S:[ S = frac{a C' e^{a t}}{1 + b C' e^{a t}} ]Now, apply the initial condition ( S(0) = 6 ). Let's plug in t = 0:[ 6 = frac{a C'}{1 + b C'} ]We can solve for ( C' ).Let me denote ( C' = C ) for simplicity.So,[ 6 = frac{a C}{1 + b C} ]Multiply both sides by ( 1 + b C ):[ 6(1 + b C) = a C ]Expand:[ 6 + 6b C = a C ]Bring all terms to one side:[ 6 = a C - 6b C = C(a - 6b) ]Solve for C:[ C = frac{6}{a - 6b} ]Given that ( a = 0.1 ) and ( b = 0.01 ):Compute ( a - 6b = 0.1 - 6*0.01 = 0.1 - 0.06 = 0.04 ).So,[ C = frac{6}{0.04} = 150 ]Therefore, the expression for S(t) is:[ S(t) = frac{a C e^{a t}}{1 + b C e^{a t}} ]Plugging in the values:( a = 0.1 ), ( C = 150 ), ( b = 0.01 ):[ S(t) = frac{0.1 * 150 e^{0.1 t}}{1 + 0.01 * 150 e^{0.1 t}} ]Simplify numerator and denominator:Numerator: 0.1 * 150 = 15, so numerator is 15 e^{0.1 t}Denominator: 1 + 0.01 * 150 = 1 + 1.5 = 2.5, so denominator is 2.5 e^{0.1 t} + 1? Wait, no.Wait, hold on. The denominator is 1 + (0.01 * 150) e^{0.1 t} = 1 + 1.5 e^{0.1 t}So, the expression is:[ S(t) = frac{15 e^{0.1 t}}{1 + 1.5 e^{0.1 t}} ]Alternatively, we can factor out e^{0.1 t} in the denominator:[ S(t) = frac{15 e^{0.1 t}}{1 + 1.5 e^{0.1 t}} = frac{15}{e^{-0.1 t} + 1.5} ]But maybe it's better to leave it as is.Alternatively, we can write it as:[ S(t) = frac{15}{1 + 1.5 e^{-0.1 t}} ]Wait, let me check:If I factor out e^{0.1 t} from the denominator:Denominator: 1 + 1.5 e^{0.1 t} = e^{0.1 t}(e^{-0.1 t} + 1.5)So,[ S(t) = frac{15 e^{0.1 t}}{e^{0.1 t}(e^{-0.1 t} + 1.5)} = frac{15}{e^{-0.1 t} + 1.5} ]Yes, that's correct.So, another form is:[ S(t) = frac{15}{1.5 + e^{-0.1 t}} ]But both forms are correct. Maybe the first form is more straightforward.Now, the question also asks for the long-term behavior as ( t ) approaches infinity.So, as ( t to infty ), let's see what happens to ( S(t) ).Looking at the expression:[ S(t) = frac{15 e^{0.1 t}}{1 + 1.5 e^{0.1 t}} ]As ( t to infty ), ( e^{0.1 t} ) grows exponentially, so the dominant terms in numerator and denominator are 15 e^{0.1 t} and 1.5 e^{0.1 t}.So, divide numerator and denominator by e^{0.1 t}:[ S(t) = frac{15}{1.5 + e^{-0.1 t}} ]As ( t to infty ), ( e^{-0.1 t} to 0 ), so:[ S(t) to frac{15}{1.5} = 10 ]Therefore, the long-term behavior is that Alex's speed approaches 10 mph.Alternatively, in the other form:[ S(t) = frac{15 e^{0.1 t}}{1 + 1.5 e^{0.1 t}} ]Divide numerator and denominator by e^{0.1 t}:[ S(t) = frac{15}{e^{-0.1 t} + 1.5} ]As ( t to infty ), ( e^{-0.1 t} to 0 ), so ( S(t) to 15 / 1.5 = 10 ). Same result.So, the speed approaches 10 mph as time goes to infinity.Let me double-check my steps.Starting from the differential equation:[ frac{dS}{dt} = aS - bS^2 ]Which is a logistic equation, so the solution should approach the carrying capacity, which in this case is ( a/b ).Given ( a = 0.1 ), ( b = 0.01 ), so ( a/b = 0.1 / 0.01 = 10 ). So, that's consistent with our result. So, the speed approaches 10 mph.Therefore, that seems correct.So, summarizing:Problem 1: The endurance function is ( E(t) = frac{5}{1 + frac{7}{3}e^{-0.2t}} ), and it takes approximately 11.17 weeks to reach 4 hours.Problem 2: The speed function is ( S(t) = frac{15 e^{0.1 t}}{1 + 1.5 e^{0.1 t}} ), and as ( t to infty ), ( S(t) to 10 ) mph.I think that's all.Final Answer1. The expression for Alex's endurance is ( boxed{E(t) = dfrac{5}{1 + dfrac{7}{3}e^{-0.2t}}} ) and it will take approximately ( boxed{11.17} ) weeks to reach 4 hours of endurance.2. The expression for Alex's speed is ( boxed{S(t) = dfrac{15e^{0.1t}}{1 + 1.5e^{0.1t}}} ) and the long-term speed as ( t ) approaches infinity is ( boxed{10} ) mph.</think>